Problem: Let $f(x)=-\dfrac{1}{2}x^4-6x^3-27x^2$. For what values of $x$ does the graph of $f$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=3$ (Choice B) B $x=-3$ (Choice C) C $x=\dfrac32$ (Choice D) D $f$ has no points of inflection.
Solution: We can find the inflection points of the graph of $f$ by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=-6(x+3)^2$. $f''(x)=0$ for $x=-3$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection point is $x=-3$. Our possible inflection points divide the number line into two intervals: $\llap{-}7$ $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $x< \llap{-}3$ $x> \llap{-}3$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<-3$ $x=-6$ $f''(-6)=-54<0$ $f$ is concave down $\cap$ $x>-3$ $x=0$ $f''(0)=-54<0$ $f$ is concave down $\cap$ We can see that the graph of $f$ does not change concavity at $x=-3$. Therefore, $x=-3$ is not a point of inflection. In conclusion, $f$ has no points of inflection.